Using the following reaction: 6 NaCl + Ba3(PO4)2 —> 2 Na3PO4 + 3BaCl2
How much barium chloride is produced (in g) when 750 g of sodium chloride react with excess barium phosphate?
The balanced chemical reaction is 3BaCl2+2Na3PO4→Ba3(PO4)2+6NaCl.
0.5 moles of barium chloride will require= "\\frac{(0.5)\\times(2)}{3} = 0.33"
However, only 0.2 moles of sodium phosphate are present.
Thus, sodium phosphate is the limiting reagent.
2 moles of sodium phosphate produce 1 mole of barium phosphate.
Thus, 0.2 moles of sodium phosphate will produce 0.1 mole of barium phosphate.
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